If cos(t) = −7/19 and tan(t) < 0, find sin(t) and cos(-t). ?

Give the answer as a fraction with a square root. Hint: Draw a triangle (perhaps within the unit circle) containing an angle with sides for the given trigonometric function.Give the answer as a fraction with a square root. Hint: Draw a triangle (perhaps within the unit circle) containing an angle with sides for the given trigonometric function.

2 Answers
May 2, 2018

sin(t)=(2sqrt78)/361
cos(-t)=-7/19

Explanation:

tan(t)<0 simply gives us quadrant information - the tangent function is negative in quadrants II and IV, while the cosine function is negative in quadrants II and III. Therefore t must occur in quadrant II.

To find sin(t), you can use the identity sin^2t+cos^2t=1. Inputting knowns:
sin^2t+(-7/19)^2=1 Simplify:
sin^2t=312/361
sin(t)=(sqrt312)/361 or (2sqrt78)/361
We know this is positive because sine is positive in QII.

Because cosine is an even function, cos(-t)=cos(t), so cos(-t)=-7/19

May 2, 2018

sint=(2sqrt78)/19and cos(-t)=-7/19

Explanation:

Here,

cost=-7/19=> color(blue)(II^(nd)Quadrant )orIII^(rd)Quadrant...to(1)

tant < 0=> color(blue)(II^(nd) Quadrant) or IV^(th)Quadrant...to(2)

From (1) and (2),we can say that

pi/2<= t <= pi to color(blue)(II^(nd) Quadrant)=>(sine) > 0

So,

(i).sin(t)=+sqrt(1-cos^2t)=sqrt(1-49/361)=sqrt312/19= (2sqrt78)/19

We know that,

"cosine is "color(red)"even function"=>cos(-theta)=costheta

(ii).cos(-t)=cost=-7/19

Note:

pi/2<= t <= pi =>-pi/2 >= (-t) >= -pi

i.e.-pi <= (-t) <= -pi/2=>III^(rd) Quadrant=> cosine < 0